Let the set of all values of \(p\), for which \(f(x)=\left(p^{2}-6 p+8\right)\left(\sin ^{2} 2 x-\cos ^{2} 2 x\right)+2(2-p) x+7\) does not have any critical point, be the interval \((\mathrm{a}, \mathrm{b})\). Then 16 ab is equal to

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Application of Derivatives

Let the set of all values of

p

, for which

f (x) = (p^{2} - 6 p + 8) (sin^{2} 2 x - cos^{2} 2 x) + 2 (2 - p) x + 7

does not have any critical point, be the interval

(a, b)

. Then 16 ab is equal to

Answer: 252

f (x) = (p^{2} - 6 p + 8) (sin^{2} 2 x - cos^{2} 2 x) + 2 (2 - p) x + 7 = - (p^{2} - 6 p + 8) cos 4 x + 2 (2 - p) x + 7

f^{'} (x) = 4 (p^{2} - 6 p + 8) sin 4 x + 2 (2 - p)

Let

k = (p - 2) (p - 4)

. For

f

to have no critical point,

f^{'} (x) \neq = 0

for all

x

, i.e.

sin 4 x \neq = \frac{2 ( p - 2 )}{4 k} = \frac{1}{2 ( p - 4 )} (p \neq = 2)

must have no solution, which requires the right side to exceed the range of

sin 4 x

in magnitude:

\frac{1}{2 ( p - 4 )} > 1 ⟹ ∣ p - 4∣ < \frac{1}{2} ⟹ p \in (\frac{7}{2}, \frac{9}{2})

(At

p = 4

f^{'} (x) = 2 (2 - 4) = - 4 \neq = 0

identically, consistent with

p = 4

lying inside this interval; at

p = 2

f^{'} (x) \equiv 0

, correctly excluded since

2 \in / (7/2, 9/2)

.)

So

(a, b) = (\frac{7}{2}, \frac{9}{2})

16 ab = 16 \times \frac{7}{2} \times \frac{9}{2} = 252

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer